Quiz 1 prep. #8-Limiting reagent  [ENDORSED]

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Quiz 1 prep. #8-Limiting reagent

Postby Priscilla_Covarrubias_HL » Tue Oct 04, 2016 9:57 pm

In question #8 in Quiz 1 Preparation, the question asks you "to determine the limiting reagent if 21.4g NH3 is reacted with 42.5g 02." I converted the 21.4g of NH3 into moles and I got 1.26 mol NH3. I also did the same for O2 and I got 1.33 mol O2. I thought that the 1.26 mol NH3 was the limiting reagent since its smaller than the 1.33 mol of O2 but then the answer key says that the limiting reagent 02. Why is that?

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Re: Quiz 1 prep. #8-Limiting reagent  [ENDORSED]

Postby Alicia_Miller_3F » Tue Oct 04, 2016 10:18 pm

Hi Priscilla,
The balanced equation for the reaction is 4NH3 + 3O2 ---------> 2N2 + 6H2O and you have to use these coefficients to determine the amount of each you would need for a complete reaction to occur and therefore which is limiting. The ratio is 4 NH3 moles for every 3 O2 moles (or 1.33). The moles you have now must be coonverted using the ratio 4:3. For example, 1.33 mol O2 x (4 mol NH3/3 mol O2)=1.77 mol NH3 and 1.26 mol NH3 x (3 mol O2/4 mol NH3)=0.945 mol O2. 1.77/0.945=1.87, a value greater than the allowed 1.33 ratio, showing that NH3 is in excess. If the final ratio value was less than 1.33, the NH3 would be the limiting reagent.

Hope this helped!

Liam Giffin 2B
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Re: Quiz 1 prep. #8-Limiting reagent

Postby Liam Giffin 2B » Thu Oct 06, 2016 12:10 am

I'm confused, because in the balanced equation you, Alicia, posted, the first product you have is 2
, but in the unbalanced equation you are given the first product is NO (g), not just N (g) or
(g). Problem 9 of the fundamentals self test in the workbook is basically the same and I was having an issue as well. It is my understanding that the balanced form of the given equation is actually 4N (g) + 5 (g) -------> 4NO (g) +6O (g).

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Re: Quiz 1 prep. #8-Limiting reagent

Postby Alicia_Miller_3F » Mon Oct 10, 2016 7:24 pm

You are right, I was going off of a different reaction equation since I did not have the workbook in front of me. However, the answer remains the same with O2 as the limiting reagent. Using 4NH3 (g) + 5O2 (g) -------> 4NO (g) +6H2O (g), the 21.4g of NH3 is converted into 1.26 mol NH3 and the 42.5g of O2 is converted into 1.33 mol O2. You can then use the coefficients to solve for the limiting reagent. For example, when converting moles of O2 into NH3, the equation is 1.33mol O2 x (4mol NH3/5mol O2) = 1.064 mol NH3. The 1.064 moles produced by O2 is less than the 1.26 moles produced by NH3 and therefore O2 is the limiting reactant.

Sorry about the mistake in equations but hopefully this clears it up for you (:

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